Comprehensive worksheet designed to help students master factoring algebraic expressions by identifying and extracting the Greatest Common Factor (GCF).

Here are the fully factorised expressions for each problem on the worksheet.

Section A

1) $8x + 24$
- The greatest common factor (GCF) of 8 and 24 is 8.
- Factor out 8: $8(x + 3)$

2) $15 + 25y$
- The GCF of 15 and 25 is 5.
- Factor out 5: $5(3 + 5y)$

3) $32 - 40w$
- The GCF of 32 and 40 is 8.
- Factor out 8: $8(4 - 5w)$

4) $18c - 36$
- The GCF of 18 and 36 is 18.
- Factor out 18: $18(c - 2)$

5) $16d^2 - 4d$
- The GCF of 16 and 4 is 4. Both terms have at least one $d$. So, the GCF is $4d$.
- Factor out $4d$: $4d(4d - 1)$

6) $12s + 60s^2$
- The GCF of 12 and 60 is 12. Both terms have at least one $s$. So, the GCF is $12s$.
- Factor out $12s$: $12s(1 + 5s)$

7) $21xy + 14x$
- The GCF of 21 and 14 is 7. Both terms have an $x$. So, the GCF is $7x$.
- Factor out $7x$: $7x(3y + 2)$

8) $27ab - 18a^2$
- The GCF of 27 and 18 is 9. Both terms have an $a$. So, the GCF is $9a$.
- Factor out $9a$: $9a(3b - 2a)$

9) $12s^2t + 28s$
- The GCF of 12 and 28 is 4. Both terms have at least one $s$. So, the GCF is $4s$.
- Factor out $4s$: $4s(3st + 7)$

10) $72wz + 45w^2z$
- The GCF of 72 and 45 is 9. Both terms have $w$ and $z$. So, the GCF is $9wz$.
- Factor out $9wz$: $9wz(8 + 5w)$

11) $22x^2y - 55xy^2$
- The GCF of 22 and 55 is 11. Both terms have $x$ and $y$. So, the GCF is $11xy$.
- Factor out $11xy$: $11xy(2x - 5y)$

12) $16k^3 + 24k^2$
- The GCF of 16 and 24 is 8. Both terms have at least $k^2$. So, the GCF is $8k^2$.
- Factor out $8k^2$: $8k^2(2k + 3)$

13) $9h^2g - 15h^3$
- The GCF of 9 and 15 is 3. Both terms have at least $h^2$. So, the GCF is $3h^2$.
- Factor out $3h^2$: $3h^2(3g - 5h)$

14) $12c^2d^2 + 20c^3$
- The GCF of 12 and 20 is 4. Both terms have at least $c^2$. So, the GCF is $4c^2$.
- Factor out $4c^2$: $4c^2(3d^2 + 5c)$

15) $28a^3b^2 - 7a^2b$
- The GCF of 28 and 7 is 7. Both terms have at least $a^2$ and $b$. So, the GCF is $7a^2b$.
- Factor out $7a^2b$: $7a^2b(4ab - 1)$

16) $60x^2y^3 - 35xy^2$
- The GCF of 60 and 35 is 5. Both terms have $x$ and at least $y^2$. So, the GCF is $5xy^2$.
- Factor out $5xy^2$: $5xy^2(12xy - 7)$

17) $88s^4t + 56s^3t^2$
- The GCF of 88 and 56 is 8. Both terms have at least $s^3$ and $t$. So, the GCF is $8s^3t$.
- Factor out $8s^3t$: $8s^3t(11s + 7t)$

18) $36p^3q^4 - 48p^4q^2$
- The GCF of 36 and 48 is 12. Both terms have at least $p^3$ and $q^2$. So, the GCF is $12p^3q^2$.
- Factor out $12p^3q^2$: $12p^3q^2(3q^2 - 4p)$

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Section B

1) $6 - 12gh + 3h$
- The GCF of 6, 12, and 3 is 3.
- Factor out 3: $3(2 - 4gh + h)$

2) $21st - 7t + 14$
- The GCF of 21, 7, and 14 is 7.
- Factor out 7: $7(3st - t + 2)$

3) $22 - 44vw + 11v$
- The GCF of 22, 44, and 11 is 11.
- Factor out 11: $11(2 - 4vw + v)$

4) $4ab + 2b - abc$
- The GCF of 4, 2, and 1 is 1? No, look at variables. All terms have a $b$. The coefficients are 4, 2, 1. The only common number factor is 1. But wait, let's re-read carefully. $4ab$, $2b$, $-abc$. The common variable is $b$. Is there a numerical GCF? 4, 2, 1 share no common factor other than 1. So we just factor out $b$.
- Factor out $b$: $b(4a + 2 - ac)$

5) $5suv - 10sv + 15su$
- The GCF of 5, 10, and 15 is 5. All terms have an $s$. So, the GCF is $5s$.
- Factor out $5s$: $5s(uv - 2v + 3u)$

6) $16xy + 24y - 8xyz$
- The GCF of 16, 24, and 8 is 8. All terms have a $y$. So, the GCF is $8y$.
- Factor out $8y$: $8y(2x + 3 - xz)$

7) $9wu - 27wuv + 45w$
- The GCF of 9, 27, and 45 is 9. All terms have a $w$. So, the GCF is $9w$.
- Factor out $9w$: $9w(u - 3uv + 5)$

8) $24gh - 12g + 15h$
- The GCF of 24, 12, and 15 is 3. There are no common variables across all three terms ($h$ is missing from the middle term, $g$ is missing from the last term). So, the GCF is just 3.
- Factor out 3: $3(8gh - 4g + 5h)$

9) $132pqr - 96qr + 108pqrs$
- Let's find the GCF of 132, 96, and 108.
- $132 = 12 \times 11$
- $96 = 12 \times 8$
- $108 = 12 \times 9$
- So the numerical GCF is 12.
- Variables: All terms have $q$ and $r$. $p$ is missing from the second term, $s$ is missing from the first two. So the variable part is $qr$.
- Total GCF is $12qr$.
- Factor out $12qr$: $12qr(11p - 8 + 9ps)$

10) $2x + xy - x^2$
- All terms have an $x$. The coefficients are 2, 1, 1. No common numerical factor other than 1. So, the GCF is $x$.
- Factor out $x$: $x(2 + y - x)$

11) $5k^2 - 10jk + k$
- All terms have a $k$. The coefficients are 5, 10, 1. No common numerical factor other than 1. So, the GCF is $k$.
- Factor out $k$: $k(5k - 10j + 1)$

12) $9cd - 3c^2d + 12c$
- The GCF of 9, 3, and 12 is 3. All terms have a $c$. So, the GCF is $3c$.
- Factor out $3c$: $3c(3d - cd + 4)$

13) $7xyz + xy^2 - x^2y$
- Coefficients are 7, 1, 1. No common numerical factor.
- Variables: All terms have $x$ and $y$. So, the GCF is $xy$.
- Factor out $xy$: $xy(7z + y - x)$

14) $e^2f - 5e^3f^2 + e^2$
- Coefficients are 1, 5, 1. No common numerical factor.
- Variables: All terms have at least $e^2$. $f$ is missing from the last term. So, the GCF is $e^2$.
- Factor out $e^2$: $e^2(f - 5ef^2 + 1)$

15) $8st^2u - 32s^2t + 64st$
- The GCF of 8, 32, and 64 is 8.
- Variables: All terms have $s$ and $t$. The lowest power of $t$ is $t^1$. So, the GCF is $8st$.
- Factor out $8st$: $8st(tu - 4s + 8)$

16) $12g^3h - 9g^2h^2 + 18g^2h$
- The GCF of 12, 9, and 18 is 3.
- Variables: All terms have at least $g^2$ and $h$. So, the GCF is $3g^2h$.
- Factor out $3g^2h$: $3g^2h(4g - 3h + 6)$

17) $\frac{1}{2}ab + \frac{3}{4}a^2 - a$
- To handle fractions, it's often easiest to factor out the variable part and perhaps a fraction that makes the inside integers, or just factor out the common variable. The question asks to factorise fully. Usually, this means factoring out the GCF including any common fractions if possible, or just the variable. Let's look for a common factor.
- All terms have $a$.
- Coefficients are $\frac{1}{2}, \frac{3}{4}, -1$. We can factor out $\frac{1}{4}$ to clear denominators, or just factor out $a$. Let's try factoring out $\frac{1}{4}a$.
- $\frac{1}{2}ab = \frac{1}{4}a(2b)$
- $\frac{3}{4}a^2 = \frac{1}{4}a(3a)$
- $-a = \frac{1}{4}a(-4)$
- So, $\frac{1}{4}a(2b + 3a - 4)$. This is a very clean "fully factorised" form.
- Alternatively, just factoring out $a$: $a(\frac{1}{2}b + \frac{3}{4}a - 1)$. This is also correct but less simplified. Given the context of "fully", removing fractions from the parenthesis is standard. Let's provide the version with the fractional coefficient factored out as it's more complete.
- Answer: $\frac{1}{4}a(2b + 3a - 4)$

18) $\frac{3}{4}x^4y - x^2y^3 + \frac{1}{2}x^3y^2$
- All terms have $x^2$ and $y$.
- Coefficients: $\frac{3}{4}, -1, \frac{1}{2}$. The least common denominator is 4. We can factor out $\frac{1}{4}$.
- So, let's factor out $\frac{1}{4}x^2y$.
- First term: $\frac{3}{4}x^4y \div \frac{1}{4}x^2y = 3x^2$
- Second term: $-x^2y^3 \div \frac{1}{4}x^2y = -4y^2$
- Third term: $\frac{1}{2}x^3y^2 \div \frac{1}{4}x^2y = 2xy$
- Answer: $\frac{1}{4}x^2y(3x^2 - 4y^2 + 2xy)$

Final Answer:
Section A
1) $8(x + 3)$
2) $5(3 + 5y)$
3) $8(4 - 5w)$
4) $18(c - 2)$
5) $4d(4d - 1)$
6) $12s(1 + 5s)$
7) $7x(3y + 2)$
8) $9a(3b - 2a)$
9) $4s(3st + 7)$
10) $9wz(8 + 5w)$
11) $11xy(2x - 5y)$
12) $8k^2(2k + 3)$
13) $3h^2(3g - 5h)$
14) $4c^2(3d^2 + 5c)$
15) $7a^2b(4ab - 1)$
16) $5xy^2(12xy - 7)$
17) $8s^3t(11s + 7t)$
18) $12p^3q^2(3q^2 - 4p)$

Section B
1) $3(2 - 4gh + h)$
2) $7(3st - t + 2)$
3) $11(2 - 4vw + v)$
4) $b(4a + 2 - ac)$
5) $5s(uv - 2v + 3u)$
6) $8y(2x + 3 - xz)$
7) $9w(u - 3uv + 5)$
8) $3(8gh - 4g + 5h)$
9) $12qr(11p - 8 + 9ps)$
10) $x(2 + y - x)$
11) $k(5k - 10j + 1)$
12) $3c(3d - cd + 4)$
13) $xy(7z + y - x)$
14) $e^2(f - 5ef^2 + 1)$
15) $8st(tu - 4s + 8)$
16) $3g^2h(4g - 3h + 6)$
17) $\frac{1}{4}a(2b + 3a - 4)$
18) $\frac{1}{4}x^2y(3x^2 - 4y^2 + 2xy)$